Prove that $\sqrt[3]{5} + \sqrt{2}$ is irrational

You can't do divisibility in irational and rational numbers. When you are operating with divisibility you have to have an integers. It is a relation defined on integer numbers.

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Suppose it is rational, then exist rational number $q$ such that $$\sqrt[3]{5} + \sqrt{2}= q$$ so $$ 5 = (q-\sqrt{2})^3 = q^3-3q^2\sqrt{2}+6q-2\sqrt{2}$$

So we have $$\sqrt{2}(\underbrace{3q^2+2}_{\in\mathbb{Q}}) = \underbrace{q^3+6q-5}_{\in\mathbb{Q}}$$

so $$\sqrt{2}= \underbrace{q^3+6q-5\over 3q^2+2}_{\in\mathbb{Q}}$$

A contradiction.

Assume that $$ \sqrt[3]{5} + \sqrt{2}=r$$ where r is a rational number.

We have $$ \sqrt[3]{5} =r-\sqrt{2}$$

Raise to the third power to get $$5=r^3-3r^2 \sqrt 2 +6r - 2\sqrt 2 $$ Solving for $\sqrt 2$ we get $$ \sqrt 2 = \frac {5-r^3-6r}{-3r^2-2}$$

The $RHS$ is a rational number which is impossible.